Find the required term of the following sequence using sequence formula: (i) 2,10,50, (6th term) (ii) 100,50,25, (7th term) (iii) 1 3 , 2 9 , 4 27 (8th term) (iv) 2,2 2, 4, (5th term)

(i) 2,10,50, (6th term) Answer: 2,10,50, T_ 6 = ? Here. a=2, r=10 2 =5, n=6 T_ 6 =a r^ n-1 T_ 6 =2 (5)^ 6-1 =2(5)^ 5 =2 3125 =6250 (ii) 100,50,25, (7th term) Answer: Here, a=100 ; r=50 100 =1 2 =n=7 T_ n =a r^ n-1 T_ 7 =100 (1 2 )^ 7-1 .=100 (1 2 )^ 6 ] =100 64 =25 16 (iii) 1 3 , 2 9 , 4 27 (8th term) Answer: Here, a=1 3 ; r= 2 9 1 3 =2 9 3 1 =2 3 ; n=8 T_ n &=a r^ n-1 T_ 8 &=1 3 (2 3 )^ 8-1 &=1 3 (2 3 )^ 7 &=1 3 128 2187 =128 6561 (iv) 2,2 2, 4, (5th term) Answer: Here. a=2 ; r= 2 2 2 =2 ; n=5 T_ n =a . r^ n-1 T_ 5 =2(2)^ 5-1 =2(2)^ 4 = (2^ 1 2 )^ 4 =2(2)^ 2 =2 4=8

Find the required term of the following sequence using sequence f…

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Question

Find the required term of the following sequence using sequence formula: (i) $2,10,50, \ldots$ (6th term) (ii) $100,50,25, \ldots$ (7th term) (iii) $\frac{1}{3}, \frac{2}{9}, \frac{4}{27} \ldots \ldots$ (8th term) (iv) $2,2 \sqrt{2}, 4, \ldots$ (5th term)

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Answer

(i) $2,10,50, \ldots$ (6th term)
Answer:
$2,10,50, \ldots \mathrm{T}_{6}=$ ?
Here. $a=2, r=\frac{10}{2}=5, n=6$
$T_{6}=a \cdot r^{n-1} $
$\therefore T_{6}=2 \cdot(5)^{6-1} $
$=2(5)^{5} $
$=2 \times 3125 $
$=6250$
(ii) $100,50,25, \ldots$ (7th term)
Answer:
Here, $a=100 ; r=\frac{50}{100}=\frac{1}{2}=n=7$
$T_{n}=a \cdot r^{n-1}$
$\therefore T_{7}=100\left(\frac{1}{2}\right)^{7-1}$
$\left.=100\left(\frac{1}{2}\right)^{6}\right] $
$=\frac{100}{64}=\frac{25}{16}$
(iii) $\frac{1}{3}, \frac{2}{9}, \frac{4}{27} \ldots \ldots$ (8th term)
Answer:
Here, $a=\frac{1}{3} ; r=\frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{9} \times \frac{3}{1}=\frac{2}{3} ; n=8$
$\begin{aligned}\mathrm{T}_{n} &=a \cdot r^{n-1} \\\therefore \mathrm{T}_{8} &=\frac{1}{3}\left(\frac{2}{3}\right)^{8-1} \\&=\frac{1}{3}\left(\frac{2}{3}\right)^{7} \\&=\frac{1}{3} \times \frac{128}{2187}=\frac{128}{6561}\end{aligned}$
(iv) $2,2 \sqrt{2}, 4, \ldots$ (5th term)
Answer:
Here. $\mathrm{a}=2 ; \mathrm{r}=\frac{2 \sqrt{2}}{2}=\sqrt{2} ; \mathrm{n}=5$
$T_{n}=a . r^{n-1}$
$\therefore T_{5}=2(\sqrt{2})^{5-1}=2(\sqrt{2})^{4}=\left(2^{\frac{1}{2}}\right)^{4}=2(2)^{2}=2 \times 4=8$